Question

(C1HW13) A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to 32.0°C from 168°C, its net radiant power decreases to 11.0 W. What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at 22.0°C as the dish cools.

Answer 1

as we know, from Stefan-Boltzmann Law,

radiation power per unit area=emissivity*stefan's constant*(T^4-Tc^4)

as emissivity and stefan's constant are constant values, let emissivity*stefan's constant=k

==>power=k*(T^4-Tc^4)

where T=temperature of the hot body

Tc=temperature of the cooler surrounding

then for T=32 degree celcius=32+273=305 K

for Tc=22 degree celcius=22+273=295 K

power is given to be 11 W

then 11=k*(305^4-295^4)

==>k=1.0182*10^(-8)

hence for T=168 degree celcius=168+273=441 K

power=k*(441^4-295^4)

==>power=308 W